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Show that cot(71∘2)=2+3+4+6 - Mathematics

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प्रश्न

Show that `cot(7 1^circ/2) = sqrt(2) + sqrt(3) + sqrt(4) + sqrt(6)`

योग

उत्तर

We have to prove that `cot(7 1^circ/2) = sqrt(2) + sqrt(3) + sqrt(4) + sqrt(6)`

L.H.S = `cot(7 1^circ/2)`

= `(cos(7 1^circ/2))/(sin(7 1^circ/2))`

To find `costheta/sintheta`, multiply numerator and denominator by 2 cos θ

Let θ = `71/2^circ`

2θ = 15°

`(2cos^2theta)/(2sin theta cos theta) = (1 + cos 2theta)/(sin 2theta)`

= `(1 + cos 15^circ)/(sin 15^circ)`

= `((1 + sqrt(3) + 1)/(2sqrt(2)))/((sqrt(3) - 1)/(2sqrt(2))`

= `(2sqrt(2) + sqrt(3) + 1)/(sqrt(3) - 1)`

Multiply numerator and denominator by `sqrt(3) + 1`

= `((2sqrt(2) + sqrt(3) + 1)(sqrt(3) + 1))/((sqrt(3) - 1)(sqrt(3) + 1))`

= `(2sqrt(2) + 3 + sqrt(3) + 1)/(3 - 1)`

= `(2sqrt(3) + 2sqrt(2) + 4)/2`

= `(2(sqrt(2) + sqrt(3) + sqrt(6) + 2))/2`

= `sqrt(2) +  sqrt(3) + sqrt(4) + sqrt(6)`

= R.H.S

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Trigonometric Functions and Their Properties
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.5 [पृष्ठ ११८]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.5 | Q 9 | पृष्ठ ११८

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